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Do check out some of puzzle solving techniques from one of best puzzle creator Cihan Altay

http://www.wpcstylepuzzles.com/2010/09/puzzle-solving-techniques-ci...

http://www.otuzoyun.com/puzzles/SolvingTechniques.pdf

This is the direct download link to the puzzle solving techniques by Cihan Altay

purifire - 2010-09-07 11:04 AM

http://www.otuzoyun.com/puzzles/SolvingTechniques.pdf

This is the direct download link to the puzzle solving techniques by Cihan Altay

Thanks for the link, rishi.

After a flop show in the Puzzle Grand prix round, I thought of attempting the left out puzzles which unfortunately are many in number.So in this Slitherlink puzzle I raced till this point and there was no trouble .But beyond this, I am clueless on how to continue through the maze.The 2 cluster looks menacing and so does the '1 area' which looks equally threatening .I can't see any way other than an endless series of guesses and contradictions to reach the end in this puzzle.What is the logical continuation?

Edited by kishy72 2014-02-07 5:05 PM

Think of the following technique.
If two one's are coming at diagonal to each other and if for one of them two the edges around connecting point is crossed then same will be reflected (mirror image) on the other 1.
Same is true if both the edges which are not originating from connecting points gets crossed.
There are lots of such points in above puzzle which will lead it to proceed in upper right part of this puzzle.

If you can able to understand and proceed using this, I will tell you one more point which will help you to proceed.

THINK IT THIS WAY..For the '1' in R3C7 the loop will either connect at the top or right of it... so continuing for the '1' at its diagonal R2C8. it will have either loop connected either at the left or down of it... if up of R3C7 is a line thn down of R2C8 wud be a line and if right of R3C7 is a line then left of R2C8 is a line. so when two such 1's are at diagonals and if either one of its opposites sides are crossed out then for the other 1 it is mirrored at its connecting point with 1... after whch, 1 at R1C9 will have a cross at it down by basic rule and even its right will be crossed out as both the ends of the loop willl go down n due to 1 at R4C10 the loop wil close..

rajeshk - 2014-02-07 9:08 PM

Think of the following technique.
If two one's are coming at diagonal to each other and if for one of them two the edges around connecting point is crossed then same will be reflected (mirror image) on the other 1.
Same is true if both the edges which are not originating from connecting points gets crossed.

Thanks for the hints fellas but unfortunately I could not make much progress with those .I haven't even managed to touch the 2 area.So a stronger trick is needed ,so that I don't need to post this puzzle again! .And thanks to this very instructive link http://logicmastersindia.com/forum/forums/thread-view.asp?tid=127&p...
which helped me start and complete the magnets puzzle just like that .

There area couple of things you can do with the 2's.
The one in the very bottom right must either be both "outside" segments or the other two,
and either way the outside edge segments NEXT to the two must be shaded in.
Now we have one segment leading into the 2 in row 8, column 10 (coming up from below).
One of the two bottom-right segments in that 2 must be included, and the other not. But
that means exactly one of the top left is included. Follow this diagonally and you can
shade in the left segment of the 2 in row 6 col 8.

There's a very nice example of this sort of thing with the 2 in r7c6, which forces
the segment coming out of the top left corner of that 2 which continues to the
left ricocheting off the O's.

Also, for example, the 2 in r6 c7 must either be the two top left or the two bottom right
segments from the x's you've put on that 1. But that means one of the two bottom
left segments of the 1 in r5c10 is included, and the right segment is not (and therefore
neither is the right segment in the 1 above it).

Those are the general sort of thing that can help you get through the thicket of 2's.

Thanks a lot! I finally completed the Slitherlink puzzle but not before having to guess twice .The above mentioned techniques played a big part in attaining the solution.I am beginning to realize that Slitherlink puzzles are very difficult,totally contrary to the earlier impression I had in mind about this puzzle genre......
I moved on to another unattempted puzzle namely the Japanese sums.The one posted below seems very different from the conventional Japanese sums puzzle in that all numbers from 1-n need not necessarily get placed in each row and column as long as the 'no number repeating in a row and column rule' is not violated.This gives the puzzle an added level of difficulty since the choices to consider for each sum are so many.So could someone please help me continue from this point?

Edited by kishy72 2014-02-12 11:19 PM

kishy72 - 2014-02-12 11:18 PM

I moved on to another unattempted puzzle namely the Japanese sums.The one posted below seems very different from the conventional Japanese sums puzzle in that all numbers from 1-n need not necessarily get placed in each row and column as long as the 'no number repeating in a row and column rule' is not violated.This gives the puzzle an added level of difficulty since the choices to consider for each sum are so many.So could someone please help me continue from this point?

Look at R7. Sum of numbers is 27. So 1 cannot be there. And the three sets must be two numbers each (27, 36, 45), which means R7C8 must be a number. This, in turn, means that R7C8, R8C8 and R9C8 must be 124. So R2C8 should be 7. and R1C8 and R1C9 should be black. You can make progress in R1 and continue.

In the 7th row, each of the three 9s have to be sum of two numbers. (2+7, 3+6, 4+5). We can deduce that the box at 7th row, 8th column contains a number. Which means that in 8th column, the second 7 = 1 + 2 + 4.
Hence, the first 7 = 7, and 3 = 3.

Other than what Rakesh and witty have said, I'll just add that your notation seems a bit counter-productive if you have to erase one of the Ns each time you fix a number. Obviously it's good to keep what suits you best. I personally circle the cells which I know will contain a number. Then I just fill in said number into the circle.

Deducing that R7C8 is a number is what I failed to notice.There was no further gimmicks in the puzzle after that and I quickly took it to completion.Thanks.....

prasanna16391 - 2014-02-13 5:20 AM

Other than what Rakesh and witty have said, I'll just add that your notation seems a bit counter-productive if you have to erase one of the Ns each time you fix a number. Obviously it's good to keep what suits you best. I personally circle the cells which I know will contain a number. Then I just fill in said number into the circle.

A very valid point that you have brought up.For one I would never use this notation in a competition scenario with time constraints.My goal here was obviously to complete the puzzle and not worry about the time that it might take to complete.I don't think that I am competent yet to complete many of the puzzle genres by myself without help or spoilers.So notation is something which looked secondary at that point to me.But I understand that notations are integral to a puzzle without which one can forget about improving and I will certainly seek to improve in this puzzle type by circling the cells which have numbers !
So continuing with the posting spree ,here is another left out puzzle from the Grand prix which despite my dedicated effort seemed far too out of reach and I am not surprised since I rarely solve a pentomino puzzle .All that I could deduce in this puzzle was that the blank cell in C4 is located somewhere between R5 and R8 and similarly the blank cell in R8 is between C1 and C4.What is the precise logical sequence that needs to be used here that makes this puzzle fall apart?!

Edited by kishy72 2014-02-14 12:07 AM

kishy72 - 2014-02-14 7:06 AM

All that I could deduce in this puzzle was that the blank cell in C4 is located somewhere between R5 and R8 and similarly the blank cell in R8 is between C1 and C4.What is the precise logical sequence that needs to be used here that makes this puzzle fall apart?!

1) You can go further with column 4. There's only one place in the column that can be empty to give 4 different pentominoes. So you can fill in all cells of column 4 at the start.
2) Your row 8 is not correct. Start with looking at that row only, you can only learn that C2 and C3 are filled. All other cells might be empty (remember there is a U-shaped pentomino for C8 and C12).
3) And your line down the bottom is too long. The central cell is filled, but you don't know yet about the cells beside it.

But the biggest hint I have for this puzzle (and other puzzle types like Camping or Battleships when missing a single clue) is that you know that the number of shaded cells must be 12x5=60, so add up the clues along the left. Whatever's left will be the clue for R7. Similarly, subtract the top clues from 60 to get the C7 clue. Easier now?

Parallel Universe II - Looks like a very tough puzzle test .Thought it would be unwise to give the test without solving the practice puzzles.So coming to the point ,can someone kindly show me the continuation in this Akichiwake Puzzle and are there any general strategies for this puzzle type?(like looking for specific clues or spots to start and proceed smoothly etc.,).It was not before spending close to 45-50 mins that I reached this stuck point....I would be glad if ideas are provided for Fill-other mino and Hypermetropic Pentominoes too.I keep breaking Fill-other mino puzzle repeatedly and the solution doesn't seem to be of much help to me in understanding how to go about the puzzle.
----Kishore----

Edited by Fred76 2014-04-25 8:35 PM

Oops, sorry !
I edited your post, thinking that it was the tournament's puzzle, and after I noticed the puzzle was from the practice material.

kishy72 - 2014-04-25 8:07 PM

Parallel Universe II - Looks like a very tough puzzle test .Thought it would be unwise to give the test without solving the practice puzzles.So coming to the point ,can someone kindly show me the continuation in this Akichiwake Puzzle and are there any general strategies for this puzzle type?(like looking for specific clues or spots to start and proceed smoothly etc.,).It was not before spending close to 45-50 mins that I reached this stuck point....I would be glad if ideas are provided for Fill-other mino and Hypermetropic Pentominoes too.I keep breaking Fill-other mino puzzle repeatedly and the solution doesn't seem to be of much help to me in understanding how to go about the puzzle.
----Kishore----

You seem to have missed the "consecutive boundaries" rule in many places. Maybe you missed this entirely while solving? Some examples - R2C2 (coz of the two boundaries to the right), R1C4, R1C6, R6C10, R7C3.

I hope you're considering the "base rules" on each page for all of "their universe" variants too. If so, and you're still getting stuck on the practice, feel free to post images. Much easier that way.

General strategy, hmm... Akichiwake is a new type. But its based on the two-consecutive boundaries rule/connected white area constraint of the Heyawake type. As of now you haven't encountered either of these in your solve yet, seemingly. Once you do, you'll see how these restrict together. E.g. - Once you shade in R6C10, you need to avoid that entire diagonally connected group of shaded cells from touching the grid edge anywhere else.

For Fill-Other-Mino,
The clues with 4 separate digits, or the edge clues with 3 separate digits are obviously going to be a polyomino of size "1". You also get 3/4 different polyominos adjacent to these clues. The rest of the steps are more complex, but just pointing out a basic start in case you've missed a rule here.

prasanna16391 - 2014-04-25 9:01 PM

You seem to have missed the "consecutive boundaries" rule in many places. Maybe you missed this entirely while solving? Some examples - R2C2 (coz of the two boundaries to the right), R1C4, R1C6, R6C10, R7C3.

Yes I missed the consecutive boundaries rule completely.Thanks very much for the ideas.I will try using them in the test.

Parallel Universe II ---That was a lovely set of puzzles.This being an open forum I will cut myself short right there.
Moving on,here are 2 puzzle genres that are present in the ongoing GP Puzzle round 4.The Prime place puzzle was something where initially I wasn't too flustered with the rules,but after seeing the queries asked in the forum of World puzzle org regarding this(and there seems to be so many) ,I am sure there won't be any who is more confused with the rules now than myself.The first is a prime place puzzle and the second is a Japanese loop puzzle where I am unable to make a start (both taken from the blog maybe puzzles).So dear LMIans, if someone has a very good idea or strategy to tackle these puzzles kindly post here.

I tried that Prime Place and broke it, so maybe I'm misunderstanding the rules. I just thought the converse part of it applies too.. I guess it does. Maybe I went wrong somewhere. But remember that by the rules, since all diagonal lines are given, all isolated cells must contain a diagonal line in the direction of their isolation (e.g. the one non-parallel line is in what I'd call an isolated cell in the direction of that line) in order to contain prime digits.

Prime Place is inherently tough. It has quite a lot of rules (I see you forgot about "connected horizontal/vertical lines must sum to a prime", which immediately deduces R1C3, R3C1, R3C5, R5C3 as primes), and a lot of bruteforcing (for example, a nonprime and a nonprime that sums to a nonprime must be both 4s like R4C1 and R5C2; also when you've identified that R1C3, R2C2, R3C1 are prime, nonprime, prime in that order, and they sum up to a prime, by bruteforce you can find that R2C2 is 1 and R1C3 and R3C1 are equal). I find a rather nice path without too much bruteforcing, which resorts to counting the 3s. Those should be sufficient hints for now?

for japanese loop i can see the start at 7 7 7.
as each needs to be two cells. there is only one way to do
then it should be good to go. :)

Invariably the following sequence happens when I see a Nurikabe puzzle in a test and it is becoming more of a nuisance to me :
1)The puzzle is attempted.
2)A lot of time is spent(wasted) on the puzzle .
3)The puzzle is left incomplete .............

This is one of the toughest puzzle genres that I have seen .The puzzle below looks very cruel and I am hopelessly lost in it.I tried to see if only certain numbers can reach some spots to avoid 2X2 oceans.There is nothing of that sort here.The shaded cell connectivity rule is also not of much help.A shaded cell maybe at R3C5 using the above rule?!Not too sure on that.So what is the approach to be used to complete this puzzle and in general what line of thinking is to be adopted while solving a Nurikabe puzzle where in most cases no logical deduction looks immediately apparent?

Edited by kishy72 2014-05-18 10:53 PM

As u Kishy72 R3C5 wil be a shaded cell... if not how wil the shaded cell at R3C6 wil join the other shaded cells.. no way for it to join other... Other thing is R1C1 and R2C1 wil be shaded as not island can go to it... Aftr which R3C2 will be a part of the island 3 or else it wil form a 2*2 shaded square. and after R3C1 wil be shaded or else the shaded part above it wil be blocked.... U can try aftr tht... if at all u r stuck u can still post.. :P

So, I was solving some Slitherlink puzzles in the hope to get in some kind of 'shape' before Puzzle Ramayan.The following is taken from Round 2 of Puzzle GP(Slovak Round).This themed slitherlink based on 1 had me clueless for a long time.After having tried several uncharted paths which got me nowhere , I tried using intuition (if I may be allowed to call what I implemented that way) : a concept which continues to baffle me and makes me wonder how people use it successfully and if it would ever help in my case , I lost hope on my futile attempts and decided to seek help here.So, if someone can show me light that would help me go until the very end,I will be thankful.

Edited by kishy72 2015-08-16 12:29 AM

Hi kishore that one even i got stuck but not at this point, I do got little progress. So I can at least tell you till that point. Bit difficult to explain but will try to keep it simple.
Notation: 9 rows and 9 columns for dots and 8 rows and 8 columns for clues. I will specify if it indicates clue and rest all will be for dots.

1) There will be line from r6c1 to r7c1 and to similar logic there will be line from r9c3 to r9c4.
2) So above step gives crosses at r5c1 to r5c2 and r5c2 to r5c3, r4c2 to r5c2 and r5c2 to r6c2.
3) Also there are crosses at r8c6 to r9c6, r4c8 to r4c9 , r3c8 to r3c9, r2c8 to r2c9, r1c8 to r2c8, r1c7 to r2c7
4) Now there will be line either from r4c2 to r4c4 or r3c3 to r5c3 to satisfy the r4c2 clue. So that gives crosses at r4c4 to r5c4 and r5c3 to r5c4.
5) Similar logic for clue at r7c2, there will be crosses r6c3 to r6c4 and r6c4 to r7c4.
6) This immediately gives crosses at, r5c4 to r6c4, r5c4 to r5c5 and r6c4 to r6c5.
7) So that gives line from r3c3 to r6c3. Also a line from r7c2 to r7c3.

Hope this helps and takes you few steps further.
If anyone has better start or logic please do let us know :)

Thanks Swaroop for explaining painstakingly and with full lucidity.These helped me a lot and I could complete the puzzle.But sadly,I could not use much of logic even after this which has to do with my own deficiency and completed it after one of my guess paths turned out to be lucky. Thanks again!

kishy72 - 2015-08-16 2:56 PM

Thanks Swaroop for explaining painstakingly and with full lucidity.These helped me a lot and I could complete the puzzle.But sadly,I could not use much of logic even after this which has to do with my own deficiency and completed it after one of my guess paths turned out to be lucky. Thanks again!

I remember that I guessed on this Slitherlink in competition and couldn't find a path later beyond a certain point either.

Could anyone give me some tips for how to approach the Fillomino Stars from the IPC last week? I found the threes in the top-left corner and the two stars in the third column, but I couldn't find anything else to do.

(Also, how do you attach a picture to a forum post here? I could upload the grid with what I have done so far.)

Edited by ghirsch 2015-08-17 12:00 AM

ghirsch - 2015-08-16 11:58 PM

(Also, how do you attach a picture to a forum post here? I could upload the grid with what I have done so far.)

http://tinypic.com/.

Upload your image.Paste the link containing the HTML version of the image here.

Thanks kishy.

This is all I could come up with for the Fillomino Stars puzzle from the IPC last week (the x's are stars). Can anyone provide some tips for where to go from here?

Edited by ghirsch 2015-08-17 1:49 AM

The top left corner cell has to be a star.

Have a marking - dot, or a small x in the corner or whatever suits you best - to denote when a cell cannot be a star, even though you don't know the region filling it. This will help. Key will be seeing R4C1 and R5C2 cannot be stars using hidden polyomino extension for contradictions.

@ghirsch
How did you mark the bottom 3-polyomino?

rakesh_rai - 2015-08-16 6:57 AM

The top left corner cell has to be a star.

Why can't it be a 1?

Rohan Rao - 2015-08-16 10:41 AM

@ghirsch
How did you mark the bottom 3-polyomino?

You're right, it could be a star.

prasanna16391 - 2015-08-16 7:06 AM

Have a marking - dot, or a small x in the corner or whatever suits you best - to denote when a cell cannot be a star, even though you don't know the region filling it. This will help. Key will be seeing R4C1 and R5C2 cannot be stars using hidden polyomino extension for contradictions.

I'll give that sort of notation a try, thanks for the advice and starting point.

ghirsch - 2015-08-17 11:14 AM

rakesh_rai - 2015-08-16 6:57 AM

The top left corner cell has to be a star.

Why can't it be a 1?

The reasoning goes this way :

i)R1 and R2 contain a total of 4 stars.
ii)Each 2X2 square can contain only a maximum of 1 star.Therefore cells R1C9,R1C10,R2C9 and R2C10 can contain a maximum of one star.Similarly Cells R1C6,R1C7,R2C6 and R2C7 can have a maximum of 1 star.
iii)To make the total 4 stars,the only place is R1C1.It has to be a star.

Thanks everyone, I figured out the rest of it now.

A NEWS Puzzle from forsmarts which I have no clue on how to go about.There is no particular reason for posting this puzzle now .I found it lying in a corner of my room along with many other unsolved puzzles.While I could do some of the rest,this caused me trouble .I had a look at the solution and still could not figure out the start.So if someone can share the starting steps ,it would be helpful.

Edited by kishy72 2015-08-20 7:34 PM

A few starting thoughts (though no major break-ins):

1) You can eliminate some spaces (the red slashes) because of the existing directions in regions (e.g. R1C6 is north of the N in the region).
2) You can eliminate some spaces (the green slashes) because the only possible entry is already in that row/column (e.g. R1C3 is northwest of the S, but the column already has an N and a W).
3) There are 22 regions (so 44 letters) and 12 rows/columns. Each type of letter can only go in 11 different rows/columns (e.g. there can't be an S in the top column), so every letter must be in 11 of 12 rows/columns.
4) This isn't included in the picture, but the region containing R5C7 can't have an N or S in it, so there must be a W in R5C7 (I'm sure this can be used elsewhere too).

Edited by ghirsch 2015-08-20 11:31 PM

Hi ghirsch. Thanks a lot for the starters.However,I am still not clear.How did you shade R2C5/C6/C7 and lots of other squares too.For instance ,why can't be a E present in R2C5 (East of South) and a W present in R1C6 and R2C6/7 (West of North)

The region containing W(R9C3) has E in R12C5 as per solution.How is it even possible?? E has to be right opposite W right?I mean in line with W.How's it located to the lower right of it or in words South East of it?

kishy72 - 2015-08-21 12:17 AM

Hi ghirsch. Thanks a lot for the starters.However,I am still not clear.How did you shade R2C5/C6/C7 and lots of other squares too.For instance ,why can't be a E present in R2C5 (East of South) and a W present in R1C6 and R2C6/7 (West of North)

The region containing W(R9C3) has E in R12C5 as per solution.How is it even possible?? E has to be right opposite W right?I mean in line with W.How's it located to the lower right of it or in words South East of it?

Hi Kishy & ghirsch,
I believe you have not interpreted rules correctly.
The alphabets has to just follow the sense of direction.
What I meant by this is, E just has to be on right hand side of W, it doesn't need to be in line with W. it can be up or down but right side.
E.g. Consider the region R8C5 "W" clue region. E can be in R8C6, R7C6 or R9C6 but it cannot be at R12C5.

Similarly for N, W , S as well.
Another example S has to be on right side and at same time downside of W.

Check this given example for clarity:
http://logicmastersindia.com/lmitests/dl.asp?attachmentid=147&view=...

Because then S won't be South of E. Look at it both ways, simply.

Thanks Swaroop and Prasanna ! I could proceed till here after which it's too difficult .Where am I missing the trick?

Edited by kishy72 2015-08-22 12:11 AM

The upside down L tetromino must have an N in the top row. Then you get an NW-pair in row 3.

rob - 2015-08-22 2:06 AM

The upside down L tetromino must have an N in the top row. Then you get an NW-pair in row 3.

I don't see how the former leads to the latter?! Where is the NW pair that you refer in row 3?How did you deduce that?

Hi kishy,
I haven't solved completely but looking at your image, I can tell you next few steps immediately:
1) R5C9 and R5C10 --> N will come in that region no matter what and in one of these places. (you can see what happens if N will not come)
2) Because of this R5C12 will be cross.
3) Now this gives N-W pair in Row 3, N or W at R3 -C5C6 AND N OR W at R3 C11-12.
4) So this eliminates N at R3 C1C2.
5) So this gives "S" at R5C1 and "E" at R3C2. So that will give "S" at r8c3.
I believe it should go easy after this.

I can't thank you enough Swaroop.I have done and dusted the puzzle finally and that's a tremendous relief after breaking my head for close to 2 days.The way you explained about that NW Pair in point 3 made me see clearly what I failed to observe....

A Numberlink puzzle.I realized that this puzzle has many ways of completing itself.Where am I going wrong?

Edited by kishy72 2015-08-30 2:50 PM

In addition,in the above puzzle,I had to use the constraint that all cells need to be visited by paths without which I found 'n' number of ways of completing this puzzle.Is there something critical that I am missing here?

kishy72 - 2015-08-30 2:49 PM

A Numberlink puzzle.I realized that this puzzle has many ways of completing itself.Where am I going wrong?

This was a broken puzzle (I suppose you have taken this from one old LMI test)

Talking about Numberlink in general, this write up (Numberlink Primer) by Palmer will be an eye-opener for most.

This was a broken puzzle (I suppose you have taken this from one old LMI test)

Ah! I see.....It was taken from Nikoli Selection 2010 LMI monthly test.

Talking about Numberlink in general, this write up (Numberlink Primer) by Palmer will be an eye-opener for most.

Thanks a lot Deb!I will look through this primer.

I ran into some problems in the following Hitori puzzle with just a few steps away from completion.Despite being so close,I fail to see a logical continuation here. Can someone post the solution ?

Edited by kishy72 2015-08-31 7:10 PM

kishy72 - 2015-08-31 6:45 PM

I ran into some problems in the following Hitori puzzle with just a few steps away from completion.Despite being so close,I fail to see a logical continuation here. Can someone post the solution ?

one quick next step: R1C2 will be shaded.
Reason: Either of R8C1 or R8C2 will be shaded so that means R7C2 is unshaded in any case. Hence above step.

swaroop2011 - 2015-08-31 8:07 PM

one quick next step: R1C2 will be shaded.
Reason: Either of R8C1 or R8C2 will be shaded so that means R7C2 is unshaded in any case. Hence above step.

Thank you ! I failed to see the 7 getting trapped.Done with the puzzle.

A very difficult Hitori Puzzle ! Can someone please post the continuation here where I am stuck?

Edited by kishy72 2015-09-02 1:43 AM

hi,
can you share the original puzzle link ?
I am confused whether what you have circled or not.
E.g. Is R6C5 "4" clue circled or not ?

swaroop2011 - 2015-09-02 2:18 AM

hi,
can you share the original puzzle link ?
I am confused whether what you have circled or not.
E.g. Is R6C5 "4" clue circled or not ?

Sorry.You are right.It's not something which I deduced logically.I forgot to erase it.

R6c5 "4:" clue should be unshaded only.

If u see all the digits above and below the numbers at R7C5/7/9/11. for each number the digits above and below is same.. so if R6C5 is shaded then for the other numbers if any of the above or
below clue is shaded the white cells get closed at top

ashaash11ash - 2015-09-02 9:18 AM

If u see all the digits above and below the numbers at R7C5/7/9/11. for each number the digits above and below is same.. so if R6C5 is shaded then for the other numbers if any of the above or
below clue is shaded the white cells get closed at top

Indeed.I missed this totally.The puzzle is still very sticky even after this deduction.But it helped a lot and I completed it .Thanks.

Just gone through the puzzle solving techniques. It's just amazing and very knowledgeable. I have learned a lot from it.

The incorrectly done Mochikoro puzzle where all islands are not connected solved in the last few mins of the test .Apparently,there are ways one can gain points by solving incorrectly too !

Edited by kishy72 2015-12-22 6:59 PM

I too gained points by solving incorrectly, in a more indirect way. I solved the Japanese Sums 25 pointer incorrectly by placing a digit by mistake where it shouldn't be, which enabled me to reach an incorrect solution and move to the 105 pointer much faster than I would have if I had solved correctly. Considering I solved the 105 pointer with 1 minute to spare, I'd say the mistake gained me 80 points

kishy72 - 2015-12-22 6:58 PM

The incorrectly done Mochikoro puzzle where all islands are not connected solved in the last few mins of the test .Apparently,there are ways one can gain points by solving incorrectly too !

prasanna16391 - 2015-12-22 11:24 PM

I too gained points by solving incorrectly, in a more indirect way. I solved the Japanese Sums 25 pointer incorrectly by placing a digit by mistake where it shouldn't be, which enabled me to reach an incorrect solution and move to the 105 pointer much faster than I would have if I had solved correctly. Considering I solved the 105 pointer with 1 minute to spare, I'd say the mistake gained me 80 points

haha interesting things.

Not sure where this question belongs, but I'm curious how much time a competitive puzzler might typically take to double-check their entries of the answer key during a timed competition. I just lost 78 points in GP Round 1 because I didn't double-check my entries. Do most people always set aside time at the end to double-check? Or after each puzzle? Or just continue to plow through the remaining puzzles and hope for the best?

For me I just plow through and hope for the best. My accuracy has improved in general with more experience though.

I could not proceed much Logical in this Fillomino Stars puzzles. Please do help me to get any logical continuation to this puzzle.

Edited by rajeshk 2016-01-31 12:17 AM

rajeshk - 2016-01-31 12:09 AM

I could not proceed much Logical in this Fillomino Stars puzzles. Please do help me to get any logical continuation to this puzzle.

You are missing a crucial deduction here just like I did when I solved it before .

1) R4C1 can't be a star .If it is a star,R3C1 can't be filled by anything.
2)Following the earlier deduction ,you have R5C2 not being a star.If it's a star,R4C2 can't take any digit 1,2 or 3.
3)Now we have R5C4 as a star as otherwise Rows 4 and 5 can't take a total of 4 stars.

Just a suggestion.Please post bigger images next time so that people find it easy to look where you are stuck : )

Edited by kishy72 2016-01-31 9:51 PM

By the way,the bottom 3's can't yet be connected(logically).This image was taken from the same stuck puzzle posted earlier in LMI.

http://logicmastersindia.com/forum/forums/thread-view.asp?tid=126&s...

Could someone point me in the right direction on Kropki #39 from the WPF GP Round 2? I'm not an expert on this puzzle type, so I'm sure I'm missing some elementary logic here, but I've gotten absolutely nowhere on this.

Encountered this frustratingly irritating Linesweeper variant that plays around with the rules in the Loop PR round.I couldn't complete it during the test (and even later), discarded it in a corner and totally forgot about it until I saw it creeping up in the LMI fun contest again.So I decided to clear up this puzzle once and for all .I continued until this point unsure of whether it is right or wrong.Could someone post the steps until the end?

What is the logic for solving this particular puzzle other than the obvious ones at the lower right and top left and the not so apparent top right ?

Edited by kishy72 2016-03-16 1:26 AM

kishy72 - 2016-03-16 1:25 AM

Encountered this frustratingly irritating Linesweeper variant that plays around with the rules in the Loop PR round.I couldn't complete it during the test (and even later), discarded it in a corner and totally forgot about it until I saw it creeping up in the LMI fun contest again.So I decided to clear up this puzzle once and for all .I continued until this point unsure of whether it is right or wrong.Could someone post the steps until the end?

What is the logic for solving this particular puzzle other than the obvious ones at the lower right and top left and the not so apparent top right ?

Firstly, your bottom left part is wrong. I don't know how you got that.
These are starting steps, which i followed. I hope it helps.

I thought the top right counts as obvious, since there is no other way to satisfy the double 4. Swaroop has the intended start.